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Mukund Rao Chaudry
Asked: 2022-11-06T02:31:38+05:30 In:

A spherical ball of mass `20kg` is stationary at the top of a hill of height `100m` , it rolls down a smooth surface to the ground , then climbs up another bill of height of `30 m` and final rolls down to a horizontal base at a height of `20 m` about the ground . The velocity attained by the ball is
A. 10 m/s
B. `10sqrt(30)m//s`
C. `40m//s`
D. `20m//s`

A spherical ball of mass `20kg` is stationary at the top of a hill of height `100m` , it rolls down a smooth surface to the ground , then climbs up another bill of height of `30 m` and final rolls down to a horizontal base at a height of `20 m` about the ground . The velocity attained by the ball is
A. 10 m/s
B. `10sqrt(30)m//s`
C. `40m//s`
D. `20m//s`
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  1. Correct Answer – C

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Pranab Hari Dhar
Asked: 2022-10-29T08:04:14+05:30 In:

A spherical ball of mass `20kg` is stationary at the top of a hill of height `100m` , it rolls down a smooth surface to the ground , then climbs up another bill of height of `30 m` and final rolls down to a horizontal base at a height of `20 m` about the ground . The velocity attained by the ball is

A spherical ball of mass `20kg` is stationary at the top of a hill of height `100m` , it rolls down a smooth surface to the ground , then climbs up another bill of height of `30 m` and final rolls down to a horizontal base at a height of `20 m` about the ground . The velocity attained by the ball is
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  1. According to conservation of energy.
    `mgH=(1)/(2) mv^(2)+mgh_(2)impliesmg(H-h_(2))=(1)/(2)mv^(2)`
    where, H=height of the first hill, `h_(1)`=height of the second hill
    `h_(2)`= height of the horizontal base
    v=velocity attained by the ball
    `implies v=sqrt(2g(100-20))=sqrt(2xx10xx80)=40 ms^(-1)`

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